. A compound contains only C, H, N, and O. It contains 37.0 % C and 42.5% O (both by mass), and there are 2 O atoms for every 1 N atom. Combustion [involving reaction with excess O2(g)] of 0.157 g of the compound produced 0.0310 g of H2O. In another, separate experiment, 0.103 g of the compound produced 0.0230 g of NH3(g). If the molar mass of the compound is 681g/mol, determine its molecular formula.

C₂₁H₁₅N₉O₁₈

Explanation:

Molecular formula is the ratio of atoms that are present in 1 molecule of the compound. We need to find moles of all atoms to find this ratio.

In a basis of 100, moles of C and O are:

Moles C:

37.0 * (1mol / 12g) = 3.083 moles C

Moles O:

42.5g * (1mol / 16) = 2.656 moles O

Now, to find moles of H, we need determine moles of H2O produced:

0.0310g H2O * (1mol / 18g) = 1.72x10⁻³ moles H2O * 2 = 3.44x10⁻³ moles H

These moles were produced when 0.157g of the compound react. In a basis of 100g:

3.44x10⁻³ moles H * (100g / 0.157g) = 2.194 moles H

In the same way, moles of N are:

0.0230g NH3 * (1mol / 17g) = 1.35x10⁻³ moles H2O

These moles were produced when 0.103g of the compound react. In a basis of 100g:

1.35x10⁻³ moles H * (100g / 0.103g) = 1.313 moles N

Empirical formula is (The simplest whole number ratio of atoms presents in a molecule). Dividing in the low number of moles (Moles N):

C: 3.083 moles C / 1.313 moles N = 2.3

O: 2.656 moles O / 1.313 moles N = 2.0

N: 1.313 moles N / 1.313 moles N = 1

H: 2.194 moles H / 1.313 moles N = 1.67

This ratio times 3 (To have the whole number ratio):

C: 7

O: 6

N: 3

H: 5

The empirical formula is:

C₇H₅N₃O₆

And weighs:

C: 7*(12g/mol)= 84

H: 5 * (1g/mol) = 5

N: 3 * (14g/mol) = 42

O: 6* (16g/mol) = 96

227g/mol

As the molecular mass of the compound is 681g/mol:

681 / 227 = 3

The empirical formula times 3 is the molecular formula, that is:

C₇H₅N₃O₆ × 3